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Homework Help: Particle in 3D box

  1. Sep 9, 2006 #1
    This is the question:

    For a particle in a three-dimensional box of sides a, b, and c, where a does not equal b and b=c, make a table of n_x, n_y, and n_z, the energies, and the degeneracies of the levels in which the quantum numbers range from 0 to 4 (Take ((a^2)/(b^2)) = 2).

    Ok, I think I have an idea of what I'm supposed to do, but I'm a little confused on two parts of the question. When they ask for a table using quantum numbers from 0 to 4, would that mean I'd have to make a long list with n_x, n_y, and n_z like:
    0 0 0
    0 0 1
    0 0 2
    0 0 3
    0 0 4
    0 0 5

    and so on until I've listed all 125? Someone in my class said that there were 64 because of the zeroes, however I didn't really understand why we would be able to. I know that:
    http://img165.imageshack.us/img165/1706/01mk3.jpg [Broken]
    So once I list all the states, I would have to substitute in the values for n_x, n_y, and n_z - but am I supposed to be able to get numerical values for the energies? I can get solve for a in terms of b and such, then plug in, but I still wouldn't get any numbers. Once I can figure out what I have to list, all I'd have to do to list the degeneracy is count the amount of states with different quantum numbers that have the same energy.
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Sep 10, 2006 #2
    I assume they mean [tex] n_x=1,2,3,4 n_y=1,2,3,4 n_z=1,2,3,4 [/tex]

    Since b=c you get 40 different cases, not 64, because for instance [tex] n_y=2,n_z=1 [/tex] and [tex] n_y=1,n_z=2 [/tex] are energy equivalent cases. You can combine [tex] n_y,n_z[/tex] in 10 different non degenerate ways(not 16) and then combine that with the 4 possible cases for [tex] n_x[/tex]

    n can not be zero thats probably what your friend mean. Consider a particle in a one dimensional box. If you put n=0 it basicly means there is no particle. the lowest state is when all three n values are =1.

    Numerical answeres doesnt seem to be possible but you can easily express them as multiples since [tex] \frac{a^2}{b^2}=2[/tex]

    Hope this helped a bit?
     
  4. Sep 10, 2006 #3
    I see what you mean now, although now I feel bad since I went ahead and listed 64 cases. At least now I can shorten my answer. Thanks a lot, this cleared up the bit about the zero.
     
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